Variational Estimates for Discrete Schr odinger Operators with - - PDF document

Variational Estimates for Discrete Schr¨

• dinger Operators

with Potentials of Indefinite Sign

David Damanik, Rowan Killip, Barry Simon, and 100DM Kick-off meeting of the EU Network Analysis & Quantum Munic, December 6-8, 2002.

1

The discrete Schr¨

• dinger operator

(Hu)(n) =

• |n−n′|=1

u(n′) + V (n)u(n) (1) with bounded potential V : Zd → R on l2(Zd). The free (discrete) Schr¨

• dinger operator, H0, cor-

responds to the case V = 0. Note that −2d ≤ H0 ≤ 2d. Remark: The discretization of the Laplacian cor- responds to H0 + 2d. Basic properties:

• σ(H0) = σess(H0) = σac(H0) = [−2d, 2d].
• If V = 0, then σ(H) = σ(H0) = [−2d, 2d].
• If lim|n|→∞ V (n) = 0 (i.e., V is compact) then

σess(H) = σess(H0) = [−2d, 2d] (Weyl’s theorem).

2

Question: Is there a converse to the above, and

if so, how does the answer depend on the dimen- sion? Motivation:

• (Killip-Simon:)

Let d = 1. If σ(H) ⊂ [−2, 2], then V = 0.

• (Rakhmanov) Let J be a general half-line Jacobi

matrix on l2(N), (Ju)(n) = anu(n + 1) + bnu(n) + an−1u(n − 1) (2) where an > 0 and N = {1, 2, . . . }. Suppose that [−2, 2] is the essential support of the a.c. part

• f the spectral measure and also the essential
• spectrum. Then limn→∞|an − 1| + |bn| = 0, that

is, J is a compact perturbation of J0, the Jacobi matrix with an = 1, bn = 0.

3

The result: There is a converse, iff d ≤ 2.

Theorem 1. Let d ≤ 2. a) If σ(H) ⊂ [−2d, 2d], then V = 0. b) If σess(H) ⊂ [−2d, 2d], then V (n) → 0 as n → ∞. (Converse to Weyl’s theorem.) Theorem 2. Let d ≥ 3. There exists a potential V with lim supn→∞ V (n) > 0 such that σess(H) = [−2d, 2d]. Remark:

• It is easy to see that for d ≥ 3 there are

potentials V ≡ 0 such that σ(H) = σ(H0) (choose V to be a small enough bump).

• Of course, if V has a fixed sign, both results

in Theorem 1 are easily proven by a simple variational argument using very simple test functions!

• The proof of Theorem 2 is done by construct-

ing a (sparse!) potential V ≥ 0 such that the Birman-Schwinger operator √ V (2d − H0)−1√ V has norm strictly less than one. This is possi- ble, since (2d − H0)−1(n, m) ∼ |n − m|−(d−2) as |n − m| → ∞.

4

More results:

• Let

d = 1. Let H be an arbitrary

• ne-

dimensional discrete Schr¨

• dinger
• perator.

Then sup σess(H) − inf σess(H) ≥ 4 with equality if and only if V (n) → V∞ a constant as |n| → ∞.

• On l2(N) (half-line): If σ(H) = [−2, 2] (i.e., no

bound states), then |V (n)| ≤ 2n−1/2

• If |V (n)| ≥ Cn−α and H has only finitely many

bound states, then α ≥ 1.

• If |V (n)| ≥ Cn−α with α = 1 and C > 1 or α < 1

and C > 0, then H has infinitely many bound states.

• Let |V (n)| ≥ Cn−α with α < 1. Then
• j

(|Ej| − 2d)γ = ∞ for γ < 1 − α 2α , where Ej are the eigenvalues of H

• utside

[−2d, 2d]. In particular, the eigenvalue sum

• j(|Ej| − 2)1/2 diverges if α < 1.

5

The variational estimate

Recall: H0u(n) =

|j|=1 u(n + j).

Then −2d ≤ H0 ≤ 2d and ϕ, (2d − H0)ϕ = 1 2

• n
• |l|=1

|ϕ(n + l) − ϕ(n)|2 Let U be given by Uϕ(n) = (−1)|n|ϕ(n). Then UH0U−1 = −H0 and UV U−1 = V. for any multiplication operator. Def: For H = H0 + V and ϕ+, ϕ− ∈ l2(Zd) set ∆(ϕ+, ϕ−) = ∆(ϕ+, ϕ−, V ) = ϕ+, (H − 2d)ϕ+ + −ϕ−, (H + 2d)ϕ−

Note:

∆(ϕ+, ϕ−) > 0 implies that H has spectrum outside [−2d, 2d].

6

Lemma 3. For f, g ∈ ℓ2(Zd) ∆(f + g, U(f − g)) ≥ 2f, (H0 − 2d)f − 8dg2 + 4Ref, V g (3) Proof. ∆(f + g, U(f − g)) = = (f + g), (H0 − 2d + V )(f + g) + (f − g), (H0 − 2d − V )(f − g) = 2f, (H0 − 2d)f + 2 g, (H0 − 2d)g

• ≥−4dg2

+4Ref, V g Remark: An obvious choice is to take f = ϕ, g = tV ϕ. The potential term on the r.h.s. of (3) is then 4t(−2dt + 1)ϕ, V 2ϕ which is maximal for t = 1

4d with 4t(−2dt + 1) = 1 2d.

7

Thus we get Lemma 4 (Comparison lemma). For any ϕ ∈ l2(Zd), ∆ ≥ 2ϕ, (H0 − 2d + 1

4d V 2)ϕ

(4) Note: The pair of test functions in this case is given by ϕ+ = ϕ(1 + 1

4dV ) and ϕ− = U(ϕ(1 − 1 4dV ))

Sometimes one would like to cut off large values

• f V (to make the norm of ϕ and ϕ± comparable).

For this one can use f = ϕ and g = tFV ϕ for some function 0 ≤ F ≤ 1. A similar calculation shows that the potential terms in the r.h.s. of (3) are given by − 8dt2ϕ, V F 2V ϕ + 4tϕ, V FV ϕ ≥ −8dt2ϕ, V FV ϕ + 4tϕ, V FV ϕ = 4t(−2dt + 1)ϕ, FV 2ϕ since F 2 ≤ F, and, again, the choice t = 1

4d is opti-

mal. So we also have Lemma 5. For any ϕ ∈ l2(Zd), 0 ≤ F ≤ 1, ∆ ≥ 2ϕ, (H0 − 2d + 1

4d FV 2)ϕ.

(5)

8

Theorem 6. Let V (n) → 0 as n → ∞. If H0 + 1

4dV 2

has at least one eigenvalue outside [−2d, 2d], then so does H0 + V . Moreover, if H0 + 1

4dV 2 has infinitely

many eigenvalues outside [−2d, 2d], then so does H0 + V .

• Proof. The first part is just the bound (4) (+ min-

max theorem), since the r.h.s. of (4) will be positive for a suitable chosen test function ϕ. For the second part one has to play with local compactness type arguments to see that one can find a sequence ϕn such that each ϕn has fi- nite support, ϕn, H0 + 1

4dV 2ϕn > 2d ϕn2, and

dist(supp(ϕl), supp(ϕm)) ≥ 2 for all l = m. This (+ Lemma 4) gives the existence of a sequence

• ϕn such that either

ϕn, (H0 + V ) ϕn > 2d ϕn2

• r

ϕn, (H0 + V ) ϕn < 2d ϕn2 and (since supp( ϕn) ⊂ supp(ϕn)) we also have ϕl, ϕm = 0 = ϕl, (H0 + V ) ϕm for l = m. Thus minmax applies again.

9

Essential spectrum and compactness:

Proposition 7 (= Theorem 1.b). Let d ≤ 2. If σess(H0 + V ) = σess, then V (n) → 0 as n → ∞. The key to this result is the fact that in dimension d ≤ 2 there are sequences ϕn such that ϕn(0) = 1 and ϕn, (2d − H0)ϕn → 0. This is not possible for d ≥ 3. Indeed, let 1 = ϕ(0) = ϕ, δ0. Since 1 = ϕ, δ0 = (2d − H0)1/2ϕ, (2d − H0)−1/2δ0 ≤

• (2d − H0)1/2ϕ
• (2d − H0)−1/2δ0
• ,

we see ϕ, (2d − H0)ϕ ≥ δ0, (2d − H0)−1δ0−1 > 0. So any ϕ ∈ l2(Zd) with ϕ(0) = 1 has a minimal kinetic energy in dimension d ≥ 3.

10

Lemma

• 8. a)

Let L1, L2 ≥ 1. There exists ϕL1,L2 ∈ l2(Z), supported in [−L1, L2], so that (i) ϕL1,L2(0) = 1 (ii) ϕL1,L2, (2−H0)ϕL1,L2 = (L1+1)−1+(L2+1)−1 (iii) for suitable constants c1 > 0 and c2 < ∞, c1(L1 + L2) ≤ ϕL1,L22 ≤ c2(L1 + L2) b) Let L ≥ 1. There exists ϕL ∈ l2(Z2) supported in {(n1, n2) | |n1| + |n2| ≤ L} so that (i) ϕL(0) = 1 (ii) 0 ≤ ϕL, (4 − H0)ϕL ≤ c[ln(L + 1)]−1 for some c > 0 (iii) (L−1 ln(L))2ϕL2 → d > 0

11

• Proof. Define

ϕL1,L2(n) =

        

1 −

n L2+1

0 ≤ n ≤ L2 + 1 1 −

|n| L1+1

0 ≤ −n ≤ L1 + 1 n ≥ L2 + 1 or n ≤ −L1 − 1 and ϕL(n1, n2) =

  

− ln[(1+|n1|+|n2|)/(L+1)] ln(L+1)

if |n1| + |n2| ≤ L if |n1| + |n2| ≥ L. These will do the job. Remark: If ψ(0) = 1 and supp(ψ) ⊂ [−L1, L2],

L2+1

• j=1

ψ(j) − ψ(j − 1) = −1 so, by Cauchy-Schwarz, 1 ≤ (L2 + 1)

L2+1

• j=1

|ψ(j) − ψ(j − 1)|2 Hence ψ, (2 − H0)ψ ≥ (L1 + 1)−1 + (L2 + 1)−1 and ϕL1,L2 is an extremal function.

12

Proof of Proposition 7 (d = 1): Assume lim sup|V (n)| = a > 0. Pick L with 2(L+1)−1 < 1

8 min(a2, 2a). Pick a sequence nj with

|V (nj)| → a and |nj − nl| ≥ 2(L + 2) for all j = l. Set F(n) = min

• 1,

2 |V (n)|

• (6)

and let ψj(n) = ϕL,L(n − nj). Then ψj, (H0 − 2 + 1

4 FV 2)ψj

≥ −2(L + 1)−1 + 1

4 F(nj)V (nj)2

≥ −1

8 min(a2, 2a) + 1 4 min(|V (nj)|2, 2|V (nj)|)

Thus we have that lim infψj, (H0 − 2 + 1

4 FV 2)ψj ≥ 1 8 min(a2, 2a)

As |FV | ≤ 2, if ϕ±,j = (1 ± 1

4 FV )ψj, we have 1 2 ψj ≤ ϕ±,j ≤ 3 2 ψj ≤ CL

(7) where CL is independent of j.

13

By Lemma 5, we have a subsequence of j’s so that either lim infϕ+,jℓ, (H0 + V − 2)ϕ+,jℓ ≥ 1

16 min(a2, 2a)

• r

lim infϕ−,jℓ, (−H0 − V − 2)ϕ+,jℓ ≥ 1

16 min(a2, 2a)

Moreover, the ϕ’s are orthogonal. Thus H has es- sential spectrum in either [2 + 1

16 C−1 L

min(a2, 2a), ∞)

• r

(−∞, −2 − 1

16 C−1 L

min(a2, 2a)] The proof of the two dimensional result is similar, using the second part of Lemma 8.

14

Proposition 9 (= Theorem 1.a). Let d ≤ 2. If σ0(H0 + V ) ⊂ [−2d, 2d], then V = 0.

• Proof. By Proposition 7, V (n) → 0. According to

Theorem 6, if H0 + V has no bound states, neither does H0 + 1

4dV 2.

Let ϕL be the function guaranteed by Lemma 8. Then 0 ≥ ϕL, (H0 + 1 4dV 2 − 2d)ϕL ≥ 1 4dV (0)2 + ϕL, (H0 − 2d)ϕL Since ϕL, (H0−2d)ϕL → 0, we must have V (0)2 =

• 0. By translation invariance, V (n) = 0 for all n.

15

Decay and bound states:

In the following we will consider the half-line oper- ator on l2(N). We will use J0 for H0 with Dirichlet b.c. at 0 (say). Proposition 10. Suppose V (n) ≥ 0 and that J0+V has no bound states. Then |V (n)| ≤ n−1 Moreover, this bound cannot be improved in that for each n0, there exists Vn0 so that Vn0(n0) = n−1 and J0 + Vn0 has no bound states. Proof: Let n0 ∈ N. J0 + δn0 has a bound state if and only if |λ| > n−1

0 . Indeed, w.l.o.g. assume λ > 0.

By Sturm oscillation theory, there is a bound state in (2, ∞) iff the solution of u(n + 1) + u(n − 1) + λδn0(n)u(n) = 2u(n) with u(0) = 0 and u(1) = 1 has a negative value for some n ∈ N. The solution is given by u(n) =

  

n n ≤ n0 n0 + (1 − λn0)(n − n0) n ≥ n0

16

Thus u gets negative iff λn0 > 1. If V (n0) > n−1

0 , then V (n) ≥ V (n0)δn0(n) for all n

(recall that V ≥ 0). By the minmax theorem and the fact that J0+V (n0)δn0 has a bound state, J0+V must have a bound state in (2, ∞). The contrapositive of V (n0) > n−1 ⇒ σ(J0 + V ) = [−2, 2] gives the first assertion of the theorem. The second follows from an explicit example.

17

Theorem 11. If J0 + V has no bound states, then |V (n)| ≤ 2n−1/2 Moreover, the above bound cannot be improved by more than a factor of 2 in that for each n0, there exists Vn0 so that J0 + Vn0 has no bound states and lim

n0→∞ n1/2

|Vn0(n0)| = 1

• Proof. If J0 + V

has no bound states so does J0 + 1

4V 2.

(By an easy extension of the compar- ison Theorem 6 to the half-line case.) By Proposition 10, we must have 1

4V 2(n) ≤ n−1.

The second claim follows from analysing the poten- tial Wn0 =

      

1 n = n0 −1 n = n0 + 1 n = n0, n0 + 1

18

The case |V (n)| ∼ βn−α:

Theorem 12. If lim infn→∞ n|V (n)| > 1, then J0+V has infinitely many eigenvalues outside [−2, 2]. Remark: We only need that for some sequence nk → ∞ 2 nk

nk

• j=nk/2

V (j)2 is not too small (that is, the potential may have zeros, but should not be sparse). If |V (n)| = βn−α and α < 1, then J0+V has infinitely many eigenvalues outside [−2, 2]. If α = 1 and β > 1, then J0 + V has infinitely many eigenvalues outside [−2, 2]. For β ∈ [−1, 1], the potential V (n) = β(−1)n−1/n has no bound states (on l2(N)).

19

A preparatory lemma: Lemma 13. For ϕ ∈ C∞

0 (R+) let

hλ[ϕ] = |ϕ′(x)|2 − λ|ϕ(x)|2 x2

• dx.

If λ > 1

4, then there exists a function ϕ ∈ C∞ 0 ((0, 1))

such that h[ϕ] < 0. More generally, let hλ,δ[ϕ] = |ϕ′(x)|2 − λ |ϕ(x)|2 (x + δ)2

• dx

for δ ≥ 0. If λ > 1

4, then there exist δ0 > 0 and ϕ ∈

C∞

0 ((0, 1)) such that hλ,δ[ϕ] < 0 for all 0 ≤ δ ≤ δ0.

Proof: The proof of the first part is just the proof

• f Hardy’s inequality, once the coupling constant is

larger than 1

4, the form is unbounded from below

(by scaling). Also by scaling, once one has a single ϕ with com- pact support for which h[ϕ] is negative, there are infinitely many of those with support in (0, 1). The second part follows from the first one by con- tinuity in δ.

20

Proof of Theorem 12: By Lemma 4, it is enough to find a sequence of functions (ϕk)k with disjoint supports, such that A[ϕk] = ϕk, (2 − H0)ϕk − 1

4ϕk, V 2ϕk < 0

(8) infinitely often. Put Uk = 2k−l, Ok = 2k+l, then Dk = Ok − Uk = (4l − 1)2n−l. Choose ε > 0 with r := lim sup(nV (n))2 − ε > 1. Furthermore, put δl =

1 4l−1. According to Lemma

13, there exist l and ϕ ∈ C∞

0 (0, 1)

such that hr/4,δl[ϕ] < 0. With ϕk(n) = ϕ(n−Uk

Dk ) = ϕ( n Dk − Uk Dk) = ϕ( n Dk − δl)

we have ϕk, (2 − H0)ϕk =

• n

|ϕk(n + 1) − ϕk(n)|2 =

• n

|ϕ(n+1

Dk ) − ϕ( n Dk)|2

= D−1

k

• n

| ϕ(n+1

Dk ) − ϕ( n Dk)

1/Dk |2D−1

k

21

and ϕk, V 2ϕk =

• n

|ϕk(n)|2V (n)2 =

• n

|ϕ( n

Dk)|2V (n + Uk)2

=

• n

|ϕ( n

Dk)|2V (Dk( n Dk + δl))2.

Since, lim infn→∞ n|V (n)| > 1, there exists K0 such that, for all k ≥ K0 and all n ≥ 0, V (Dk( n

Dk + δl))2 ≥

r (Dk( n

Dk + δl))2 = D−2 k

r ( n

Dk + δl)2,

in particular, ϕk, V 2ϕk ≥ D−1

k

• n

|ϕ( n

Dk)|2

r ( n

Dk + δl)2D−1 k

. for all large enough k.

22

Since

• n

| ϕ(n+1

Dk ) − ϕ( n Dk)

1/Dk |2D−1

k

→

• |ϕ′(x)|2 dx

and

• n

|ϕ( n

Dk)|2

r ( n

Dk + δl)2D−1 k

→ r

• |ϕ(x)|2

(x + δl)2 as k → ∞, we see that lim

k→∞ DkA[ϕk] =

|ϕ′(x)|2 − r

4

|ϕ(x)|2 (x + δ)2

• dx

= hr/4,δl[ϕ] < 0. To finish the proof, one only has to note that we can choose a subsequence kj such that dist(supp(ϕl), supp(ϕm)) ≥ 2 for all l = m .

23

Blowup of suitable eigenvalue moments: Lemma 14. Let |V | ≤ 4d on supp(ϕ). Then there exists ψ with supp(ψ) = supp(ϕ) so that ψ−2

• ψ, (H0 + V )ψ
• − 2d

≥ 1

4

• ϕ−2ϕ, (H0 + 1

4d V 2)ϕ − 2d

• Proof. Put ψ± = (1 ± (4d)−1)ϕ.

By assumption, ψ± ≤ 4 ϕ. The claim follows from the basi- Comparison Lemma 4, choosing either ψ+ or ψ− for ψ. To convert this into bounds for eigenvalue mo- ments we use the following elementary Lemma 15. Let A be a bounded selfadjoint opera-

• tor. Let {ϕj}∞

j=1 be an orthonormal set with

ϕj, Aϕk = αjδjk If F is a nonnegative even function on R that is monotone nondecreasing on [0, ∞), then tr

• F(A)
• ≥
• j

F(αj)

24

Theorem 16. Let J be a Jacobi matrix of the form J0 + V where |V (n)| ≥ Cn−α for some α < 1 and V (n) → 0. Then

• j
• |Ej| − 2

γ = ∞

for γ < 1 − α 2α where Ej are eigenvalues of J outside [−2, 2]. Remark: If the constant C is large enough, one gets divergence for γ < 1−α

2α .

An inspection of the proof shows that this result is independent of the dimension! (Replacing |Ej| − 2 by |Ej| − 2d, of course.)

25

• Proof. Fix p > 0. Let ϕm be supported near mp+1
• n an interval [mp+1 − C1mp, mp+1 + C1mp] (C1 is

picked to arrange that the supports are separated by at least 2). Taking the slopes fixed on each half-interval, we see ϕm, (2 − H0)ϕm ≤ C2 mp (9) ϕm, 1

4 V 2ϕm ≥

C3mp m2α(p+1) (10) and ϕm, ϕm ≥ C4mp. So long as α(p + 1) < p (that is, p <

α 1−α), (10)

beats (9) for large m, and we find ϕm, ϕm−1ϕm, (H0 + 1

4 V 2 − 2)ϕm ≥ C5m−2α(p+1)

Note that, as p ↓

α 1−α, 2α(p + 1) ↓ 2α 1−α.

By the last lemma with F(x) = dist(x, [−2, 2])γ, we see that we have divergence if γ < 1−α

2α .

26